City R and City S is 375 km apart. Henry left City R for City S at 7.00 a.m. travelling at an average speed of 75 km/h. Wesley left City R later than Henry and caught up with him at 10.00 a.m. Wesley was travelling at a speed of 100 km/h.
- At what time did Wesley leave City R?
- How much later did Henry arrive in City S than Wesley? Express your answer in h as a fraction or mixed number.
(a)
From 7.00 a.m. to 10.00 a.m. = 3 h
Distance that Henry covered when Wesley caught up with Henry
= 3 x 75
= 225 km
Time that Wesley took to travel 225 km
= 225 ÷ 100
= 2
14 h
1 h = 60 min
14 h =
14 x 60 = 15 min
2
14 h = 2 h 15 min
Time that Wesley left City R:
2 h 15 min before 10.00 a.m. = 7.45 a.m.
(b)
Remaining distance that they had to cover to reach City S
= 375 - 225
= 150 km
Duration that Wesley had to travel before reaching City S
= 150 ÷ 100
= 1
12 h
Duration that Henry had to travel before reaching City S
= 150 ÷ 75
= 2 h
Duration that Henry took to arrive later than Wesley in City S
= 2 - 1
12=
12 h
Answer(s): (a) 7.45 a.m.; (b)
12 h
